Fluid - Work done to blow a soap bubble

Hi everybody, I now encounter some problems when try to solve this:Problem statement: Calculate the work done to blow a soap bubble to radius R. Knowing that the process is isotherm, the atmospheric pressure is H, the surface tension is γ.

Solution:$$A = A_{1}+A_{2}$$
In which A1 is the work done to blow the bubble with 2 layers :$$A_{1} = 2(4 \gamma πR^{2}) $$
A2 is the work done to blow the bubble to pressure p' = H + 4γ/R
$$A_{2}=p' V ln \dfrac{p'}{H}$$

The solution above is in my textbook, but I still help some questions
First, why the soap bubble has 2 layers, which leads to the Laplace pressure is 4γ/R?
Secondly, in A2 equation, why the ratio is natural logarithm is p' over H? In the isotherm, we have: $$\dfrac{V_{i}}{V_{f}}=\dfrac{p'}{H}$$
It doesn't match the formular used to calculate work done in the isotherm process.

Secondly, in A2 equation, why the ratio is natural logarithm is p' over H? In the isotherm, we have: $$\dfrac{V_{i}}{V_{f}}=\dfrac{p'}{H}$$
It doesn't match the formular used to calculate work done in the isotherm process.

It should match. Can you state the formula that you have in mind for the work done in an isothermal process?